Determine

Question

Determine $\lim_{x\rightarrow\infty}f\left(x\right)$ and $\lim_{x\rightarrow-\infty}f\left(x\right)$ for the following functions. Then give the horizontal asymptotes of $f$ (if any).

$f (x) = \frac{1}{2 x^{4} - \sqrt{4 x^{8} - 9 x^{4}}}$

Accepted Answer

Welcome back. Everyone. In this problem, we want to evaluate the following limits and determine the horizontal asymptotes. If any for the function F FX equals one divided by three X squared minus the square root of nine X to the fourth minus 16 X squared. For our answer choices A says the limit as X approaches infinity of F FX is 3/8 as it approaches negative infinity is also 3/8. And the horizontal Asymptote is 3/8 B says it's 3/8 negative 3/8. And the horizontal asymptotes are 3/8 and negative 3/8 C says its infinity, negative infinity and no horizontal Asymptote. And D says all of them are zero. Now, if we're going to find the following limits, let's try to make our, our function a bit easier to work with. OK. Now, for our feng Shan, OK, if I fix which is equal to one divided by three X squared minus the square root of nine X to the fourth minus 16 X squared, the first thing we can do is to do, sorry to make this function simpler is to multiply both the numerator and denominator by the conjugate of the denominator and that will get the square root out for a denominator. So we can multiply it by three X squared plus the square root of nine X to the fourth minus 16 X squared in the numerator. And by the same expression in our denominator. Now, when we do that, OK, since our numerate for the numerator of F FX is one, OK, then we know we're gonna have three X squared plus the square root of nine X to the fourth minus 16 X squared as our numerator. But in our denominator, when we multiply both terms, we end up is the difference of two squares. So we end up with nine X to the fourth minus nine X to the fourth minus 16 X squared. Now we can even simplify that further in our denominator. OK? To get our denominator to be equal to oops, let me just clean that up here to be equal to positive 16 X squared. No, we can simplify F FX even further because no, we can divide every term by X square. OK. So dividing through, dividing through by X squared. OK. Keeping in mind that the square root of X to the fourth is just another way to write X squared. OK. Then F of X, we can write F of X as three X squared divided by X squared plus the square root of nine X to the fourth divided by X to the fourth because it's under the square root minus 16 X squared divided by X to the fourth. OK. And all of that is being divided by 16 X squared divided by X squared. Now, in this case, X squared cancels X squared X to the fourth, cancels X to the fourth X squared divided by X to the fourth equals one divided by X squared. OK. And in our denominator X squared would also cancel X squared. So now we can rewrite F of X, OK. As three plus the square root of nine minus 16 divided by X squared. OK. All divided by 16. Now that we have it in a simpler form, we can evaluate our limits much easier. OK. So no, OK. This means that the limit as X approaches infinity of F FX is going to be equal to the limit as X approaches infinity of three plus the square root of nine minus 16 divided by X squared divided by 60. Now, in this problem here, we have an X term for 16 divided by X squared. That's the only place where we have an X squared term. And as X approaches infinity, that means 16 divided by X squared inside the square root approaches zero. So that means our limit then is going to be equal to three plus the square root of nine divided by 16. OK. The square root of nine is three. So that's gonna be three plus three or six divided by 16, which equals 3/8. OK. Now that's our limit as X approaches infinity. What about the limit as X approaches negative infinity? Well, now in this case, again, as X approaches negative infinity, the term inside the square root is also going to approach zero. So we are also going to end up with three plus the square root of nine divided by 16, which we've seen eventually equals 3/8. Now, since both limits as X approaches infinity and negative infinity are equal to 3/8 that means the given function has a horizontal Asymptote at Y equals 3/8. OK. So F FX has a horizontal asy to at why equals 38. Therefore, the limit as X approaches infinity as it approaches negative infinity and its horizontal Asymptote equals 3/8. If we look back at our answer choices that tells us a is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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